Question: Mario programs his helicopter to fly at a target with a velocity (speed and direction) of $\vec p$. The direction of $\vec p$ is due east, and its speed is $6\,\text{m/s}$. To Mario's dismay, the wind causes his helicopter to actually fly with a velocity $\vec a$, and it misses its target. The direction of $\vec a$ is $30^\circ$ south of east, and its speed is $5\,\text{m/s}$. (Assume "due east" is $0^\circ$, "due north" is $90^\circ$, and so on.) What is the speed of the wind?
Answer: Consider vector $\vec w$ (depicted below), which represents the wind. We can imagine how it would cause Mario's helicopter to slow down and change direction. It is reasonable to assume that the velocity of the wind added with the programmed velocity of Mario's helicopter equals the resultant (actual) velocity of his helicopter. $\vec w + \vec{p} = \vec{a}$ Before finding $\vec w$, we'll need to split both vectors into their components. $\vec p$ is $6\,\text{m/s}$ due east, which can be written as $(6, 0)$. $\vec a$ can be broken into its components using trigonometry. (Angle not drawn to scale) $x = 5\cos{(-30^\circ)} = 5\dfrac{\sqrt 3}{2} \approx 4.330$ $y = 5\sin{(-30^\circ)} = -\dfrac{5}{2} = -2.5$ Therefore, $\vec a = (4.330,\, -2.5).$ We can now solve for $\vec w$. $\begin{aligned} \vec w + \vec p &= \vec a\\\\ \vec w &= \vec a - \vec p\\\\ \vec w &= (4.330,\, -2.5) - (6,0)\\\\ \vec w &= (-1.670, -2.5) \end{aligned}$ We can find the magnitude of $\vec w$ (i.e., the speed of the wind) using the Pythagorean theorem. $\begin{aligned} \| \vec w \|^2 &= (-1.670)^2 + (-2.5)^2\\\\ \| \vec w \| &= \sqrt{(-1.670)^2 + (-2.5)^2}\\\\ \| \vec w \| &\approx 3.0 \text{ m}/\text{s} \end{aligned}$ $\vec w$ is pointing into the third quadrant, so we can find its direction (call it $\theta~$ ) using the arctangent function and adding $180^\circ$. $\begin{aligned} \tan \theta &= \dfrac{-2.5}{-1.670}\\\\ \theta &= \arctan{ \left ( \dfrac{2.5}{1.670} \right ) } \\\\ \theta&\approx 56^\circ \end{aligned}$ Adding $180^\circ$ to this result gives us the actual direction, $236^\circ$ (rounded to the nearest degree). The speed of the wind is $3.0 \text{ m}/\text{s}$. The direction of the wind is $236^\circ$.